 Renewal process

A renewal process counts successive events whose interarrival times ${\xi }_1,{\xi }_2,\dots$ are independent and identically distributed. Allow nonnegative finite interarrival times, but require $\mathbb{P}({\xi }_1>0)>0$. Its renewal epochs and count areThe nondegeneracy condition makes $S_n\to \infty$ almost surely, so the count is finite at every finite time. Zero interarrival times can give simultaneous renewals; strictly positive ones give distinct epochs. At each renewal the future interarrival sequence starts afresh, which models repeated replacement or completion events.

Exponential interarrival times give a Poisson process. Constant interarrival times ${\xi }_i=c>0$ give $N_t=\lfloor t/c\rfloor$. A general renewal count is not a Markov process by itself: predicting the next arrival usually requires knowing the time since the last renewal, because a general interarrival law is not memoryless.

If $0<\mu =\mathbb{E}{\xi }_1<\infty$, then $N_t/t\to 1/\mu$ almost surely. The strong law gives $S_n/n\to \mu$; sandwiching $t$ between $S_{N_t}$ and $S_{N_t+1}$ proves the counting-rate limit. No exponential or continuous interarrival distribution is required for this long-run rate.